WAEC Further Mathematics Past Questions & Answers - Page 54

266.

Two bodies of masses 8 kg and 5 kg travelling in the same direction with speeds x m/s and 2 m/s respectively collide. If after collision, they move together with a speed of 3.85 m/s, find, correct to the nearest whole number, the value of x.

A.

2

B.

5

C.

8

D.

13

Correct answer is B

\((m_{1} v_{1} + m_{2} v_{2}) = (m_{1} + m_{2})v\) (Inelastic momentum)

\(8x + (5 \times 2) = (8 + 5) \times 3.85\)

\(8x + 10 = 13 \times 3.85 = 50.05\)

\(8x = 50.05 - 10 = 40.05\)

\(x \approxeq 5 m/s\)

267.

Find the area of the circle whose equation is given as \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)

A.

\(3\pi\)

B.

\(6\pi\)

C.

\(9\pi\)

D.

\(12\pi\)

Correct answer is C

Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Given that \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)

Expanding the equation of a circle, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Comparing this expansion with the given equation, we have

\(2a = 4 \implies a = 2\)

\(-2b = 8 \implies b = -4\)

\(r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9\)

\(r = 3\)

\(Area = \pi r^{2} = \pi \times 3^{2}\)

= \(9\pi\)

268.

Two vectors m and n are defined by \(m = 3i + 4j\) and \(n = 2i - j\). Find the angle between m and n.

A.

97.9°

B.

79.7°

C.

63.4°

D.

36.4°

Correct answer is B

\(m . n = |m||n|\cos \theta\)

\((3i + 4j) . (2i - j) = 6 - 4 = 2\)

\(2 = |(3i + 4j)||(2i - j)| \cos \theta\)

\(|3i + 4j| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)

\(|2i - j| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\)

\(2 = 5(\sqrt{5})(\cos \theta)\)

\(\cos \theta = \frac{2}{5\sqrt{5}} = 0.08\sqrt{5}\) 

\(\theta = \cos^{-1} 0.1788 = 79.7°\)

269.

Given that \(\log_{2} y^{\frac{1}{2}} = \log_{5} 125\), find the value of y

A.

16

B.

25

C.

36

D.

64

Correct answer is D

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 125\)

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 5^{3} = 3\log_{5} 5 = 3\)

\(\log_{2} y^{\frac{1}{2}} = 3 \implies y^{\frac{1}{2}} = 2^{3} = 8\)

\(y = 8^{2} = 64\)

270.

The tangent to the curve \(y = 4x^{3} + kx^{2} - 6x + 4\) at the point P(1, m) is parallel to the x- axis, where k and m are constants. Determine the coordinates of P.

A.

(1, 2)

B.

(1, 1)

C.

(1, -1)

D.

(1, -2)

Correct answer is C

\(y = 4x^{3} + kx^{2} - 6x + 4\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6\)

At P(1, m)

\(\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0\) (parallel to the x- axis)

\(6 + 2k = 0 \implies k = -3\)

\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)

= -1

P = (1, -1)